By T. Peterfalvi, R. Sandling

The recognized theorem of W. Feit and J. G. Thompson states that each team of wierd order is solvable, and the facts of this has approximately elements. the 1st seemed in Bender and Glauberman's neighborhood research for the strange Order Theorem, quantity 188 during this sequence. the current e-book presents the character-theoretic moment half and completes the facts. Thomas Peterfalvi additionally deals a revision of a theorem of Suzuki on cut up BN-pairs of rank one, a prerequisite for the type of finite easy teams.

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9) Theorem. 6). Let k be such that 0 < k < and let T = {/lj 10 < j < W2, /lj(l) = W2 /lk(l)}. (a) If /lj E T, then /lj E T and /lj =I- /lj. Also, 0=1- Z[T, L#] = Z[T, A]. (b) The Z-linear mapping from Z[T] to Z[Irr G] which sends the character /lj to Ok LOSi

Let Tl be an isometry from Z[Sl] to Z[Irr G] which extends the restriction ofT to Z[Sl' L#]. Set (Xi -aiXlr = ~() (Xl(1)Xi Xl 1 xi(l)xd T ; this is compatible with previous notation if ai E N. 1) Let (X - aXlt = X - Y, where X E Z[R(X)] and Y is orthogonal to R(X). There is an integer A E Z such that n Y _ TI - aXl - A ' " ai ~ -II '112 XiTI ,=1 X, Z + , where Z E CF( G) is orthogonal to S{'. Proof. Set Y = ax? - 2:;'=1 AiX? + Z with Ai E C and where Z E CF(G) is orthogonal to S{'. For 1 ::; i ::; n, X?

32, it follows that Irr(K) has at most W2 elements left fixed by g. But, since {lOj E Irr(L), Xj is fixed by g. It follows that, if X E Irr(K) is not one of the characters Xj, then X is not fixed by g. b), Ind~ X is irreducible. Moreover, (Ind~ X, {lij) = (X, Xj) = O. Finally, if {l E Irr( L) and if X is an irreducible component of Res~ {l, then {l is a component of Ind~. X, and so is one of the characters {lij or is of the form Ind~ X. 6) Hypothesis. 2). 1). (c) Let H be a normal subgroup of L such that W 2 c H C f{.